Question

An air chamber of volume ‘V’ and pressure ‘P’ has a neck area of cross-section ‘A’ into which a ball of mass ‘m’ just fits and can move up and down without friction. When the ball is pressed down a little and released it executes S.H.M. The time period of oscillations of the ball is (assuming pressure – volume variations to be isothermal). ($p_0$ = atm pressure)

• A. $2\pi \sqrt{\frac{mV}{PA}}$
• B. $\frac{2\pi }{A}\sqrt{\frac{mV}{P}}$
• C. $2\pi \sqrt{\frac{P_{0}mV}{PA}}$
• D. $\frac{2\pi }{A}\sqrt{\frac{mV}{P+P_0}}$

Answer 1

The correct option is B

$\frac{2\pi }{A}\sqrt{\frac{mV}{P}}$

$pA-p_{o}A=mg$

After compression pressure changes to $p_1$

$pv=p_1(v-Adx)$ or $p_1=\frac{pv}{(V-Adx)}$

$f_{restoring}=p_{1}A-mg+p_{0}A$

$ma=(p_1-p)A$ or $ma=p\left(\frac{Adx}{V-Adx}\right)$ A or $ma=\frac{P{A}^{2}dx}{m(V-Adx)}\Rightarrow a=\frac{P{A}^2}{mV}dx$ Assuming ‘dx’ to be very small)

or $\omega =\sqrt{\frac{P{A}^2}{mV}}\Rightarrow T=\frac{2\pi }{A}\sqrt{\frac{mV}{P}}$